Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:

/**
 * DA-IICT
 * Author : Savaliya Sagar
 */
import java.io.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;
import static java.lang.Integer.*;

public class C816 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	void solve() {
		int n = ni();
		int m = ni();
		int g[][] = new int[n][m];
		int r[] = new int[n];
		int c[] = new int[m];
		Pair p[] = new Pair[n * m];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				g[i][j] = ni();
				p[m * i + j] = new Pair(i, j, g[i][j]);
			}
		}
		sort(p);
		int f = 0;
		int total = 0;
		for (int i = 0; i < n * m; i++) {
			int x = p[i].u;
			int y = p[i].v;
			if (n < m) {
				int r1 = checkr(g, n, m, x);
				r[x] += r1;
				redr(g, n, m, x, r1);
				int c1 = checkc(g, n, m, y);
				c[y] += c1;
				redc(g, n, m, y, c1);
				total += r1 + c1;
			} else {
				int c1 = checkc(g, n, m, y);
				c[y] += c1;
				redc(g, n, m, y, c1);
				int r1 = checkr(g, n, m, x);
				r[x] += r1;
				redr(g, n, m, x, r1);
				total += r1 + c1;
			}
			if (g[x][y] > 0) {
				f = 1;
				break;
			}
		}
		if (f == 1) {
			out.println(-1);
		} else {
			out.println(total);
			for (int i = 0; i < n; i++) {
				while (r[i]-- > 0) {
					out.println("row " + (i + 1));
				}
			}
			for (int i = 0; i < m; i++) {
				while (c[i]-- > 0) {
					out.println("col " + (i + 1));
				}
			}
		}
	}

	int checkr(int[][] g, int n, int m, int i) {
		int min = MAX_VALUE;
		for (int j = 0; j < m; j++) {
			min = Math.min(min, g[i][j]);
		}
		return min;
	}

	int checkc(int[][] g, int n, int m, int j) {
		int min = MAX_VALUE;
		for (int i = 0; i < n; i++) {
			min = Math.min(min, g[i][j]);
		}
		return min;
	}

	void redr(int[][] g, int n, int m, int i, int k) {
		for (int j = 0; j < m; j++) {
			g[i][j] -= k;
		}
	}

	void redc(int[][] g, int n, int m, int j, int k) {
		for (int i = 0; i < n; i++) {
			g[i][j] -= k;
		}
	}

	class Pair implements Comparable<Pair> {
		int u;
		int v;
		int g;

		public Pair(int u, int v, int g) {
			this.u = u;
			this.v = v;
			this.g = g;
		}

		public int hashCode() {
			int hu = (int) (u ^ (u >>> 32));
			int hv = (int) (v ^ (v >>> 32));
			return 31 * hu + hv;
		}

		public boolean equals(Object o) {
			Pair other = (Pair) o;
			return u == other.u && v == other.v;
		}

		public int compareTo(Pair other) {
			return -1 * compare(g, other.g);
		}

		public String toString() {
			return "[u=" + u + ", v=" + v + "]";
		}
	}

	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Thread(null, new Runnable() {
			public void run() {
				try {
					new C816().run();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b !=
									// ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n && !(isSpaceChar(b))) {
			buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}
}